Trace of product of two Hermitian matrices

Question

Let $A$ and $B$ be two Hermitian complex matrices. (a) Prove that $\operatorname{tr}(AB)$ is real. (b) Prove that if $A, B$ are positive, then $\operatorname{tr}(AB)>0$.

(a) The trace of Hermitian matrix is a real number, since $a_{ii} = \bar{a}_{ii}$, that also means that all eigenvalues are real. I can't proceed to conclusion that $\operatorname{tr}(AB)$ is real, since $\operatorname{tr}(AB) \neq \operatorname{tr}A\cdot\operatorname{tr}B$ and product of two Hermitian matrices is also Hermitian only if these matrices commute, which is not the case for arbitrary Hermitian matrices.

(b) Am I missing something or the question is indeed so easy? If all entries $a_{ij}>0$ and $b_{ij}>0$, then all $c_{ij}=\sum_{m}a_{im}\cdot b_{mj}>0$ too, finally $\operatorname{tr}(AB)=\sum_{i}c_{ii}>0$.

Answer 1

Since $A^T = \overline{A}$ and using basic properties of trace :

$$ tr(AB) = tr((AB)^T) = tr(B^T A^T) = tr(\overline{B} \ \overline{A})= tr(\overline{A} \ \overline{B}) $$

Therefore :

$$ tr(AB) = tr(\overline{A} \ \overline{B}) = tr(\overline{AB}) = \overline{tr(AB)} $$

Finally $tr(AB) \in \mathbb R$.

Answer 2

You can't just say that because the trace is over the diagonal elements of $(AB)_{ii}$. Instead $\lambda = tr(AB)=\sum A_{ij}B_{ji}$ then $\lambda^* = tr(A^*B^*)=tr(A^TB^T)=\sum A^T_{ij}B^T_{ji} = \sum A_{ji}B_{ji}$ which is clearly $tr(AB)$ under renaming dummy indicies, so $\lambda=\lambda^*$